Question

In the given figure, CM and RN are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR, then prove that Δ CMB ~ Δ RNQ.

Answer 1

Given: Δ ABC ∼ Δ PQR

Their corresponding sides are proportional, and their corresponding angles are equal: 

∴ ∠ A = ∠ P, ∠ B = ∠ = Q, ∠ C = ∠ R    ...(1)

`(AB)/(PQ) = (BC)/(QR) = (AC)/(PR)`   ...(2)

CM and RN are medians of ΔABC and ΔPQR respectively.

In Δ CMB ∼ Δ RNQ

∠ B = ∠ Q    ...(Included angle is equal)

`(AB)/(PQ) = (BC)/(QR)`    ...[From equation (2)]

`(2 MB)/(2 NQ) = (BC)/(QR)`

`(MB)/(NQ) = (BC)/(QR)`    ...(Sides are proportional)

Since two sides are proportional and the included angle is equal, the triangles are similar by the SAS (Side-Angle-Side) similarity criterion. 

Δ CMB ∼ Δ RNQ

Hence Proved.